15th - 1 1.7 P( )VVICR SERIES . _ 1; 11—1 _ 2. lnn...

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Unformatted text preview: 1 1.7 P( )VVICR SERIES . _ 1; 11—1 _ 2. lnn U“—“ < 1 2:» lnn (x l < 1 5‘ < 1 6 < X < 4', when X = —(3 we have 11 —1 oc- Us 11 —1 oc- (X1 .1“ TX! 7:: z (—1)“, a divergent series; when X = —4 we have 2: 1, a divergent series not 1171 (a) the radius is 1; the interval of convergence is —6 < X < —4 (b) the interval of absolute convergence is 76 < X < 74 (c) there are no values for which the series converges conditionally - 11.,+ - (x—2]"_l 1on x—2 5.1111111” u—"J<1=>nlll]}x__l W'W 2:» —8 < X < 12; when X = —8 we have (—1)“, a divergent series; when X = 12 we have 1, a divergent 11—1 11—] series (a) the radius is 10; the interval of convergence is —3 < X < 12 (b) the interval of absolute convergence is —3 < X < 12 (c) there are no values for which the series convelges conditionally 1) - uni“ - 10'“ 7 ntlnnli - r1 - lnn ‘ 27' llllpbc- 1.1.. <1 2:} 111—19139 (n11](lntn1]1_‘1"‘ K" < l 2:} lxl (nl—lpbc- n11) lntntlJ) < l 1. 3 2 =>|X|(1) lim_ (1’) <1=>|X|(li1n_m1) <1=>|X|<1=>—1<X<l;whenX=—lwehave [l—‘CXJ 11—‘09 n TI: TI: (—1 1" - _. ,. _ _. 1 - . E l wlnch conv erges absolutel) , when X — 1 we ha\ e E m Wthll converges n_ n—l (a) the radius is 1; the interval of convergence is 71 < X < 1 (b) the interval of absolute convergence is —1 < X < 1 (c) there are no values for which the series convelges conditionally 2n+3 x— n 32. Hut “1—” <1 2:» li1n 145] 11—100 Us 11—100 ( I) 2 _ 2 . ::>(x.—:/—]<1::>(X—\,/§) <2::> X—\/§<\/§:¢>— 6<X—\/§<\/§:¢>0<X<2\/§;when 211— x’ ‘1» 2 3° :1 1 2 3° _ , , . . X = Owe have 2: $ = — Z 2 g“ = — Z we which diverges s1nce “11111)C an $ 0; when X = EVE n—1 n—1 n—l ‘ I 2n+l so ,5 so so \, . n 12 . . we have 2: ( 2]" = Z 3 g“ = 2: V6, a d1vergent series n—l n—l 11—] (a) the radius is V5; the interval of convergence is 0 < .X < 2V5 (b) the interval of absolute convergence is 0 < X < 2V5 (c) there are no values for which the series convelges conditionally . ml 2 s - ., 37.nli‘mxl “:1: <1 => “11ng (“13”) 1%)“ <1=>("3”-’n1_11nw|1|<1=> *3” <1 => x~<2 2:» < x/E 2:» — G < .X < x/E; at X = ; VG we have (1)rl which diverges; the interval of convergence is n—Cl x. _l n — 6 < X < x/E ; the series 2: (k 3' 1) is a convelgent geometric series when —\,/E < .X < V5 and its sum is 1-1—0 11.8 TAYLOR AND MACIAURIN SERIES 3. fix) 2 3? = f], f’(x) = —x_2, f"(x) 2 211—3, f’"(:x) = —6x_4; f(2) , f’(2) = — ET, f”(2) = 1:, f’”(x) = — =>P.,(x)=%,P](x)=%—}T(x—2),P3(x)=%—3T(x—2)—%(x—2)', P;1(x)=?3 11x 2) 11x 213 3,11: 21" 6. f(x)=cosx,f’(x)=—sinx,f"(x)=—cosx,f’”(x)=sinx;f(§]=eosgzfi, PG] :—sin%:—7]E,f”(%] :—e05§:—7]E,f’”(%] :sinfizfi 2:» P1100: V12, P100=fig-figb-fiththfi-figfi-fl-fi§(x-%Ji 2 :1 flatware—452%) —,3—51x—11 12. fix): (1 —x)_] => fix): (1 —x)‘3.f”(x) = 211— 111—3, 1%) = 3:11 — 111—4 => 1mm 2 ma — 11-11-2110) 2 1, 1'10) 2 1, 9’10) 2 2, 1111(0) = 3:, ,fmi-(m = k! '35 =1—x—x2—X“—... =2: x“ :8 1—K n-D fix) 2 3x“ — x4 — 2.313 — x2 — 2 2:» f’(x) = 15.31“1 — 4x3 — 6x2 — 2x, f"(x) = 60113 — 12.313 — 12x— 2, 1mm 2 180112 — 24.x — 12,1001“: 3601: — 24, £15101) 2 360, Mm = 0 ifn :-~ 6; f(— 1) = —7, 91—11 = 23,1"1—1): —82,f”’(—1) = 216, 1(411—1) = —334, 1(5?(— 1) = 360, 1031—1) 2 Oifn :5 6 24. =1» 3.1“ x4 2113 x3 2 = 7 23(11 1) 3%: 1)2 231,601 11" 343,411: 1)4 355,001 1)“ = 7 231’): 1) 41(11 1].2 36(x 1)?‘ 1601 1)“1 3(11 1)“ 11.9 CONVERGENCE OE TAYLOR SERIES; ERROR liS'l‘IPB/IA'IES . L2 2" _ _ 1:211 ; 113 35 1—D" _ n31: 6- = 1111;: j (5—,) = ((1%) ) = 2;] 1 = z 1']— Fl— n— : 1 _ _ x_“ _ x_"' _ 2-2I 21-4! 24-6! 15 If; : Ken—121]: K3 Z;an : Zoznxnm : X3 _ 2x3. _23K4 _ 23.x". _ 22 V l — x = 1 — 5, — g: — — . By the Alternating Series Estimation Theorem the |err01'| < % < @ = 1.25 X 10—“ 33. [5111'] 1: when x = %;the sum is tan—1 :3 0.808448 11.10 APPLICA'I‘IONS OE POWER SERIES ._--) . _1 3 1’ _L _2 _§':1" , 7- [l_x‘..,]1;-=l Fifi ( .,](2!]1x) (£11 1;!( 111x) :1 13x1 gxh 156x: 13. (1 — 211)3 = 1 31 2x) (MEL—2""! 13““131'2” = 1 6x 12112 3114* 24. [1— x3] 3/ — 211)? = a] (2513 2am): (3613 251] 511))? (4:14 2:13 233))(3 (narl nan_g)x“ = 0 2:» a] = 0, 2:13 — 2a“ 2 0, 333 — 351] = 0, 4:14 — 4513 = 0 and in general naJ1 — nan_2 = 0. Since y = 3 when x: 0,we]1a\-'e a“ = 3. Therefore :13 = —3,ay, 20,34 = 3, ,Elgm] 20,512rl = (—1)“3 ,_ 2 .1 _ 3° 2 _ 3° 2 n_ 3 2:» }—3—3x —3x —... —Z%)3(—l)“x“—Z%3[—x] —m_r A 3 _1_ 31. }”—x3y=233 633): (4-33.; am}? (n(n Dan an_4)x“‘2— . 2x 2:» 233:0,63321, 4 - 334 — an 2 0, 5 - 435 — a] = 0, and in general n(n — Dan — an_4 = 0. Since 3/ = b and y = a when x = CI, wehavean : and a] = b. Therefore :12 = 0,02. = = — fins : — 4%“ -. 20,617 = if“ , _ 1 :1. a 4 b r. 1 7 2m" bx" 2* .‘ — a b" 2-3 X 3-4 x 4-5 X 2-3-07 9‘ 3 4 s 4 5-3-9 F(X):J:([2_%_%_%_HI)dt:[In 1|] 1L:- ;. _ :» |e1'1'01'| < :5 0.000013 54.0: l)sin(x]}]]=(x 1)(X1H 1 1 .‘,)=1_ 1 20m 11:1 S!(x} 11-1 you 1:? _ sum 11' _ ::>xlip})cl(x—l)5in(x:]]: lim_l (1 1 1 .mjzl ...
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15th - 1 1.7 P( )VVICR SERIES . _ 1; 11—1 _ 2. lnn...

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